# Proof of Gauss's Lemma (Riemannian Geometry version)

I was self-learning Do Carmo's Riemannian Geometry, there is a step in the proof of Gauss's Lemma what I can't quite figure out.

Since $d\,\exp_p$ is linear and, by the definition of $\exp_p$, $$\langle (d\,\exp_p)_v(v),(d\,\exp_p)_v(w_T)\rangle=\langle v,w_T\rangle.$$

So I went on wikipedia hoping to find something that can help me figure this out.

I did find something. HERE. It says that $(d\,\exp_p)_v(v)=v$. In order to do that, it constructs that curve $\alpha(t)$ with $\alpha(0)=v$, $\alpha'(0)=v$. And it gives $\alpha(t)=(t+1)v$. I agree with all these. Then it argues that you can view $\alpha (t)=vt$ since it's just a shift of parametrization. Okay. I am okay with that. But in this case, should $(d\,\exp_p)_v(v)={d\over dt}(\exp_po \alpha(t))|_{t=1}$, instead of evaluating the derivative at $t=0$ as argued in wiki?

Also, I found myself really uncomfortable with all the abuse of notations in Differential Geometry. Like here, the identification of the $T_p M$ and $T_v(T_p M)$ freaks me out. Does this mean that when a local coordinate system is picked, $T_p M$ under the natural basis will be the same as $R^n$, and so does $T_v (T_p M)$?

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$T_p M$ is a real vector space, and in any real vector space $V$ (with a chosen basis or not!) there is a canonical linear isomorphism $V \cong T_v V$, for all "points" $v$ in $V$. There's nothing to freak out about at all here! – Zhen Lin Jul 27 '12 at 5:52
@ZhenLin Thank you for the clarification! – henryforever14 Jul 27 '12 at 8:08

The equation you are citing in combination with the reasoning you are citing, too, is used in do Carmo only for $w_T$ parallel to $v$. Otherwise the equation would be still correct (that is the content of the Gauss lemma) but requires a different justification (as presented later on in Do Carmo).
If $w_T$ is parallel to $v$ then $\exp_p(tw)$ is just a parametrization of the geodesic through $p$ in direction $v$ with constant speed $|w|$ (this is where the definition of $\exp$ is used) and the derivative in your formula can be computed as the derivative of this geodesic w.r.t to $t$, so it is simply the tangent vector to that geodesic of length $|w|$ in the point under examination.
I assume you mean $\exp_p (tw_T)$. But I would like to know why the method used in Wiki is also right. By the definition of the differential of a map, to compute $(d\,\exp_p)_v(v)$, one construct a curve $\alpha(t)$, with $\alpha(0)=v$(the point where the differential is evaluated) and $\alpha'(0)=v$ (the direction). So a candidate for such $\alpha(t)$ would be $\alpha(t)=(t+1)v$. Now, by definition, $$(d\,\exp_p)_v(v)={d\over dt}(\exp_po \alpha(t))|_{t=0}={d\over dt}\gamma(1,p,(t+1)v)|_{t=0}={d\over dt}\gamma(t+1,p,v)|_{t=0}$$And then, how does it conclude that the result is $v$? – henryforever14 Jul 27 '12 at 7:59
I think the result should be the parallel transport of $v$ along the geodesic at the point $\exp_p(v)$. This makes sense, and it yields what I need. Because Parallel transportation preserves the inner product if the connection is compatible with the metric. – henryforever14 Jul 27 '12 at 10:10
@henryforever14 The $v$ was a typo, thanks for pointing that out. The curve $\exp_p \circ\alpha$ is just the geodesic I mentioned, by definition of $\exp_p$. Therefore the result of the differential in direction $v$ is the tangent to the geodesic in the point under consideration, this is what you wrote down in your comment. This is not $v$, but the parallel transport of $v$, since, yes, the tangent of a constant speed geodesic is parallel along the geodesic. I don't know whether the reasoning in Wikipedia is correct. – user20266 Jul 27 '12 at 10:16